My team recently hired someone. Normally, this wouldn't be such a big deal, but we've
been looking for someone for a year and a half. In this time, we've interviewed at least a
dozen candidates and phone-screened at least a couple dozen more. Practically every
candidate we've talked to had at least two years of C++ experience, rated himself a 7 or 8
on a scale of 10 in C++ skill, and had one or two lucrative offers on the table.
Unfortunately, we would have rated almost all of them somewhere between a 4 and a 6. In my
opinion, this goes to show you that working with C++ for a long time doesn't guarantee you
really understand the language.

Over this time, I've developed a stock interview question that's proven to be a pretty
good gauge of C++ knowledge. No one has yet been able to just rip out the correct answer,
but we've had several, including the guy we hired, who understood the important issues and
were able to get the question right with prompting. As a public service, I'd like to share
my stock question and its answer with you and explore the various programming issues it
presents.

You have a class, TFoo, which descends from a class, TSuperFoo, and
which has two data members, both of which are pointers to objects of class TBar.
For the purposes of this exercise, consider both pointers to have owning semantics and TBar
to be a monomorphic class. Write the assignment operator for this class.

This seems like a simple enough exercise, but it gets at some interesting issues. It's
a good way to test a programmer's grasp of C++ syntax and C++ style, but more importantly,
it tests the programmer's knowledge of C++ memory management and exception handling.

For the impatient among you, let's cut right to the chase: One correct answer to this
question would look something like this:[*]

Yes, it's a lot of code. Yes, it's ugly. But all the code you see here is necessary.
We'll go through it all piece by piece and see why this is.

"But I never have to write an assignment operator!"

The first reaction I usually get from people is something along the lines of "But
I never have to write assignment operators." You should. If you've ever created a new
class, you've needed to write an assignment operator.

Let's examine why this is so. In C++, there are three things every object is expected
to be able to do: An object should be able to initialize itself to a default state, it
should be able to initialize itself from another instance of the same class, and it should
be able to assume the semantic state of another instance of the same class. In C++, these
operations are expressed with the default constructor (e.g., TFoo::TFoo()), the
copy constructor (TFoo::TFoo(const TFoo&)), and the assignment operator (TFoo::operator=(const
TFoo&)).

These three functions are special in C++: If you don't provide them yourself, C++
provides them for you. And automatically makes them public. Among other things, this means
you have to define these operations even if you don't want a client to be able to
copy or default-construct a particular class. If you don't want a class to be copied, for
example, you have to define an empty copy constructor and assignment operator yourself and
make them private or protected.

Furthermore, the compiler isn't guaranteed to create versions of these classes that do
exactly what you want them to do. For copying and assignment, for example, the
automatically-generated code will do a shallow memberwise copy. If your class has
pointer members, this is practically never what you want, and even when you don't have
pointer members, this isn't always the right behavior. It's definitely not what we want in
our example.

Even when the default versions of the special functions do what you want them to, it's
still generally a good policy to always spell that out explicitly by writing them
yourself. It avoids ambiguity, and it forces you to think more about what's going on
inside your class. Always give any new class a default constructor, a copy constructor,
and an assignment operator.

Copy vs. assign

Another misconception I see often is a fuzzy idea of the difference between the copy
constructor and the assignment operator. They're not the same thing, although they're
similar. Let's take a moment to look at the difference.

The copy constructor and assignment operator do similar things. They both copy state
from one object to another, leaving them with equivalent semantic state. In other words,
both objects will behave the same way and return the same results when their methods are
called. If they have public data members (generally a bad idea), they have the same
values. This doesn't necessarily mean that the objects are identical: some purely internal
data members (such as caches) might not be copied, or data members pointing to other
objects might end up pointing to different objects that are themselves semantically
equivalent, rather than pointing to the same objects.

The difference between the copy constructor and assignment operator is that the copy
constructor is a constructor  a function whose job it is to turn raw storage
into an object of a specific class. An assignment operator, on the other hand, copies
state between two existing objects. In other words, an assignment operator has to
take into account the current state of the object when copying the other object's state
into it. The copy constructor is creating a new object from raw storage and knows it's
writing over garbage. For many classes, the current state of the object doesn't matter and
both functions do the same thing. But for some classes (including the one in our example),
the current state does matter, and the assignment operator is more complicated.

Defining the Function

What parameters does the function take? C++ requires that an assignment operator
take one parameter: the thing on the right-hand side of the = sign. This can be of any
type, but the assignment operator that C++ automatically generates for you (and therefore,
the one we're interested in here) is the one where you have the same type of object on
both sides of the = sign. That means the parameter is either an instance of or a reference
to an instance of the same class as the object on the left-hand side. You'll pretty much
always want to use a reference rather than a full-blown instance of the class (i.e., pass
by reference instead of pass by value). This is because passing an object by value
requires creating a new instance of the class with the same state as the object passed as
a parameter: in other words, its copy constructor must be called. This isn't necessary,
and it wastes time. The parameter can be either a const or a non-const reference, but
since it would be terrible form for the assignment operator to have side effects on the
object on the right-hand side, you should use a const reference.

What does the function return? An assignment operator can return anything it
wants, but the standard C and C++ assignment operators return a reference to the left-hand
operand. This allows you to chain assignments together like so:

x = y = z = 3;

Unless you have a really good reason, you want to follow this convention.
Returning a reference to the right-hand operand or a value (i.e., another whole TFoo)
would both still allow the simple chain described above to work, but have subtle
differences in semantics from the way the standard operators do it that would come out in
more complicated expressions involving assignment. Returning a value also forces
unnecessary trips through the object's copy constructor, costing you in performance.

So the outer shell of a properly-written assignment operator would look like this:

(By the way, note that the return statement is "return *this",
rather than "return this". That's because this is a pointer
to TFoo. You have to dereference it to get a reference to TFoo. Of
course, if you forget this, the compiler will remind you.)

Virtual or non-virtual? I had one applicant suggest that operator=
should be a virtual function. Let's take a look at this issue. Many C++ programmers are
trained to make everything virtual, and in fact, some older frameworks do just that. In
the specific example of the assignment operator, however, it's not a good idea. An
override of a virtual function has to take the same parameters as the function it's
overriding. Therefore, TFoo's operator= function would have to be
declared as

virtual TSuperFoo& TFoo::operator=(TSuperFoo& that);

You could declare the function this way and still have it return this, of
course, because a reference to a TFoois a reference to a TSuperFoo,
but we have no way of knowing whether that is a reference to a TFoo or a
reference to some other subclass of TSuperFoo. If it's not a TFoo, you
have several problems. You'd have to check that's class, which can be expensive.
If it isn't a TFoo, you obviously wouldn't want to try to carry out the
assignment, but then you'd have to define some kind of error-handling protocol to handle
this situation. Better just to make operator= take the right type and let the
compiler check the classes of your operands for you at compile time.

Of course, as soon as each class has operands with different types, the functions have
different signatures and the operator= function is no longer being overridden. So
it doesn't make sense to make operator= virtual.

Owning and Aliasing Pointers

Okay, now that we've got the preliminaries out of the way, we can get into the actual
nitty-gritty of having our assignment operator actually perform an assignment. Let's
refresh our memory of what the object we're working on looks like:

Unfortunately, that's the wrong answer here. Remember that in the original question I
said that fBar1 and fBar2 are owning pointers. To understand why
the above example won't do what we want, we need to take a look at the unique problems of
C++ memory management.

Because of its evolution from C, C++ is much closer to the hardware than most other
object-oriented languages. One of the chief consequences of this is that you have to do
your own memory management. Every new that happens during a program's execution
must be balanced by one and only one delete. You don't want objects you've
allocated to clutter up memory after you're done with them, you don't want to try to
delete an object more than once and you don't want to access an object after you've
deleted it. Double-deleting an object can corrupt the memory manager's free list, leading
to crashes down the road; reading through a pointer to a deleted object (a "dangling
pointer") can lead to wrong results; and writing through a dangling pointer can
corrupt other objects or cause crashes. Failing to delete an object you're done with (a
"memory leak") is less obviously malignant, but can seriously degrade
performance and eventually cause crashes when the system runs out of memory.

In a system of any complexity, sticking to this "one delete for every new"
rule can be quite difficult, so a strict protocol for managing memory is necessary. The
basic rule we follow at Taligent is that for every object in the runtime environment,
there is one and only one pointer to it through which the object can be deleted. This
pointer is an "owning pointer," and the object or function containing that
pointer is the object's "owner." All other pointers to the object are called
"aliasing pointers." The owner of the object expects to delete the object;
objects with aliasing pointers don't.

So when we say that TFoo's two TBar pointers are owning pointers,
we're saying that TFoo expects to delete those TBars. In other words,
its destructor looks like this:

TFoo::~TFoo()
{
delete fBar1;
delete fBar2;
}

You can see that if more than one TFoo object points to a given TBar,
then as soon as one of those TFoos is deleted (taking the TBars down
with it), the other TFoos are hosed. The next time any one of them tried to
access one of its TBar objects, it'd be reading or writing through a dangling
pointer, with potentially disastrous consequences. Therefore, every TFoo must
have its own unique TBar objects, which means our assignment operator must create
new copies of the source object's TBars for the destination object to point to.

In some cases, of course, it's overkill to make a copy of an object, because the
current owner of the object is just going to delete that object after passing its content
on to another object. In other words, one object is transferring ownership of an
object to another object. This happens quite frequently, in fact. A simple factory method
starts out with ownership of the object it creates, but when it returns its value, it
passes ownership of that object to the caller. Its return value is an owning pointer. At
other times, a function returns a pointer to an object but intends that the caller merely
use it for a short time to perform some operation. Ownership is not transferred; the
return value is an aliasing pointer.

When you have functions that return pointers or have pointer parameters, you must make
it explicit whether the function transfers ownership, and you must then make sure that
code calling the function upholds these semantics. C++ doesn't do any of this for you.
Sometimes you can do this through the parameter types (references are virtually always
aliases, and const pointers are always aliases), and sometimes you have to do it with
naming conventions (at Taligent, for example, we use "adopt,"
"orphan," and "create" in the names of functions that transfer
ownership).

In the case of the assignment operator, our parameter is a const reference to another TFoo.
That alone signifies that we are not taking ownership of its internal state (there are
some very rare, but important, exceptions to this rulewe'll look at one later).
Since TFoo's pointers are defined as owning pointers, however, we have to
reconcile the difference in semantics by making new copies of the objects the other TFoo
points to.

As an aside, it's worth mentioning that there are occasions where copying objects to
maintain ownership semantics is too expensive. In these situations, a technique called reference
counting can be used. We'll touch briefly on reference counting later.

memcpy() is evil!

So we have to copy the TBars in our assignment operator. I've seen a lot of
interesting attempts to do this. The most frightening was

This would copy the data members of that's TBars into this's
TBars, so this and that retain their own separate TBar
objects. So we're doing well so far. The problem is that it bypasses the assignment
operator and copy constructor for TBar, so if TBar has any owning
pointers of its own, you have the same problem. You'll also have problems if TBar
owns locks or system resources that need to be properly cleaned up or duplicated when you
change the internal state of the object. And, of course, if any of these pointers is NULL,
you'll probably crash.

This is kind of a cheap way of initializing brand-new TBars from existing
ones, or copy constructing them without using the copy constructor. It suffers from all of
the same limitations as the previous example, plus an additional one we'll get to in a
moment.

Keep in mind one thing: memcpy() is evil! It's a C construct you should
never use in C++. memcpy() operates on bits and bytes, not on objects. At best,
it just looks ugly and forces you to be concerned with things you shouldn't need to worry
about, like the size of TBar. At worst, it fails to take into account what's
actually being stored in the objects you're copying, leading to erroneous results, or even
uglier code that takes the special cases into account. Never use memcpy() in C++
code. There are always better, more object-oriented ways to do the same thing.

So what's the right answer? Since the TFoo expects to delete the TBars
it points to, we have to create new ones for it to point to. And we can create duplicates
of the other TFoo's TBars by using TBar's copy constructor
(remember from our introduction that every object has a copy constructor), so the correct
solution (so far) would look like this:

Plugging the leaks

Of course, there's still a glaring error here: Remember that fBar1 and fBar2
are owning pointers. This means that TFoo is responsible for deleting them. Here,
we've copied right over the top of these pointers without taking into account their former
values. This'd be okay if we were writing a copy constructor, where we're guaranteed that fBar1
and fBar2 contain garbage, but it's not okay for an assignment operator. In the
assignment operator, fBar1 and fBar2 are both valid pointers to TBar
objects. If you just write over them, you now have two TBar objects in memory
that nobody points to anymore (or at least, no one who points to them expects to
have to delete them). This is a memory leak. Memory leaks won't cause your program to
crash or produce wrong results, at least not right away. Instead, depending on how
numerous and bad they are, they'll slowly degrade your program's performance. And if you
run the program long enough, you'll run out of memory and it will crash.

So we have to delete the objects that we currently own before we can create new ones:

Assigning yourself

Now we're beginning to get to something reasonable. But we're not quite there yet.
Consider the following expression:

foo1 = foo1;

This might seem like kind of a silly example, but it does happen. Consider a setter
function on some object where the variable being set is a TFoo and the other
value is passed in. A caller knows he wants to set that variable to "x." The
caller shouldn't have to check to see whether the value of that variable is already
"x." If the value already is "x" and the caller doesn't check
for this, look at what happens in our code: this and that refer to the
same object, so by the time we get down to "fBar1 = new TBar(*that.fBar1)",
that.fBar1 is gone. "delete fBar1" also deleted that.fBar1.
The call to TBar's copy constructor will either crash because it's trying to
access a deleted object, or it'll get away with that, create a brand-new TBar,
and initialize it with the potentially random contents of raw memory. Worse, most of the
time the data that had been in those two objects won't have been overwritten yet, so it'll
probably work right 90% of the time and randomly fail the other 10%. This kind of bug is
notoriously hard to track down.

There are many ways of coding around this, but the obvious answer is the best: just
check at the top of the function to see whether you're assigning to yourself, and drop out
if you are.

Of course, you can't just write

if (this == that)

because this and that are different types (this is a pointer
and that is a reference). There are two ways of rectifying this: you can turn this
into a reference, or turn that into a pointer. In other words, you could write

if (*this == that)

or you could write

if (this == &that)

Keep in mind that these two expressions don't do the same thing. The first example
tests semantic equality, and the second example tests identity. In other words, "if
(*this == that)" calls TFoo's equality operator. If that object doesn't
have one (it's not required or created for you), you're automatically hosed. If it
does, it'll go through all the externally-visible data members of the two objects one by
one looking for a difference, which is bound to be much slower than "if (this !=
&that)", which is a simple pointer comparison. "if (this ==
&that)" will return true only if this and that refer to
the same object, not merely to two objects that "look the same." You'll
occasionally do some unnecessary work this way, but it saves a lot of time and it protects
you from crashing, which is really the point.

Honoring your ancestry

We've also forgotten one other important detail. Remember the first line of our sample
class definition:

class TFoo : public TSuperFoo {

TFoo is not a root class; it has a base class. This means we have to copy over
the base class's data members too. If we were writing a copy constructor, we wouldn't
generally have to worry about this, because the compiler will make sure our base class
members are initialized before our constructor is called. But the compiler doesn't do
anything like this for us with assignment operators; we have to do it ourselves.

The easiest way to do this is to call our superclass's assignment operator ourselves.
You could do this by casting yourself to your base class:

*((TSuperFoo*)this) = that;

but it's much more readable to just call the inherited function by name:

TSuperFoo::operator=(that);

By the way, I've had a couple of people try

inherited::operator=(that);

Some of the older compilers on the Mac provided the inherited keyword, which I
always liked. Unfortunately, it's not included in the C++ standard because it doesn't work
well with multiple inheritance: in portable C++, you actually have to refer to your
immediate superclass by name. So now we have

Cleaning up after yourself

We're still not really out of the woods here. The code above will work great... unless
we encounter an error while trying to create one of our TBar objects. If we get
an exception while creating a TBar object, the data member we're setting retains
its old value, which now points to a deleted object. If we continue to use this TFoo,
we'll probably eventually crash because of the dangling pointer. If we delete the TFoo
in response to the exception, we'll probably blow sky high trying to double-delete the TBar.

If we know that the calling function will delete the object if assigning to it fails,
we can just zero out the pointers after deleting the objects they refer to:

Unfortunately, this only works if you're certain the calling function will catch the
exception and delete the TFoo, or if NULL is a valid value for the TBar
pointers. (Actually, if NULL is a valid value, you'll also have to check the
incoming object's fBar1 and fBar2 for NULL before trying to new
up new TBars from them.)

A better way to handle the problem is to make sure that the creation of a new object
succeeds before you do anything to the variable you're assigning to. This way, if creation
fails, you're still pointing to a perfectly good TBar the assignment
operation simply didn't have an effect. In fact, since we have two TBars, we
should new up both of them before carrying out the assignment. This will ensure
that the TFoo is always in an internally consistent state; either the whole
assignment happened or none of it did:

But there's a problem with this solution: Consider what happens if creation of bar1
succeeds and creation of bar2 fails. You'll exit with the actual object
untouched, but what happens to the TBar pointed to by bar1? That's
right; it leaks. In order to avoid this, you actually have to catch and re-throw the
exception and delete bar1 if it's been created. In order to tell if you're
created bar1, you need to set it to NULL first, too. And, so
we're really sure the assignment doesn't do anything until we know we've been able to
create both of the new TBars, we can't call our inherited operator=
function until after the try/catch block. So this all gets rather complicated:

Delegating our work

Of course, we're not really handling the exception here; we're just catching the
exception to enable us to clean up properly. The try/catch block is a really ugly
construct to have to use in this way. It'd be really nice if we could lose it.

One of the niceties of C++ exception handling is that destructors for any stack-based
objects are guaranteed to be called even if the function that declares them exits
prematurely with an exception or a return statement. We can take advantage of this
behavior by having an object whose destructor deletes the object that would otherwise
leak.

The new ANSI C++ standard provides us with just such an object: It's called auto_ptr.
Using auto_ptr, we can write

The release() function gets rid of the auto_ptr's reference to the
object so that it won't delete the object in its destructor. So the auto_ptr will
only delete the object it points to if we exit the function with an exception before
getting to the release() call. Which is correct; the only things in this function
which can fail are the constructor calls.

Taking full advantage

You've probably already guessed this part: We can actually utilize auto_ptr
more fully than this. auto_ptr actually implements owning pointer semantics for
us; if you assign to an auto_ptr, it deletes the object it points to before
taking on the new pointer, and if an auto_ptr goes out of scope, it also
automatically deletes the object it points to. So if we relaxed the rules of our exercise
to allow us to redefine the class, we could redefine the class using auto_ptr:

Functions accessing the objects pointed to by fBar1 and fBar2 would
look exactly the same as they did when fBar1 and fBar2 were regular
pointers; auto_ptr defines its * and -> operators to do the
same thing as those for a regular pointer.

And now we can take full advantage of auto_ptr in our assignment operator:

auto_ptr also takes care of deleting the objects previously pointed to by fBar1
and fBar2. One thing it doesn't automatically do for us, however, is create a new
object to point to; we have to do that ourselves. If we just did

fBar1 = that.fBar1;
fBar2 = that.fBar2;

instead, we'd actually inadvertently affect that, when that's not what we want
to do (in fact, by declaring that as const, we've promised not to affect
it). This is because an auto_ptr's assignment operator passes ownership from one
object to the other (this is the once exception to the rule about passing ownership that I
mentioned earlier); in other words, it'd leave that.fBar1 and that.fBar2
with null pointers.

Actually, our auto_ptr solution doesn't do exactly the same thing as the
previous example. If we want to make sure that all of the assignment happens or none of
it, we still need the temporary variables. But this is a situation where the
ownership-passing property of auto_ptr's assignment operator helps us:

Here, if the second new operation fails, the first new TBar will be
deleted by auto_ptr's destructor when we exit the function. But if both news
succeed, the assignments will delete the objects fBar1 and fBar2
previously pointed to, and will also zero out bar1 and bar2 so that
their destructors don't delete anything when we exit the function.

The other beauty of auto_ptr is its documentary value; if a pointer is
declared as an auto_ptr, you know it's an owning pointer. If you consistently use
auto_ptrs for all owning pointers and regular pointers only for aliasing
pointers, the meanings of pointers are no longer ambiguous and you don't have to worry as
much about naming conventions and documentation.

Of course, auto_ptr isn't available on all C++ compilers yet; if you're
concerned about portability, don't use it. Do the assignment the first way I described, or
make your own auto_ptr-like class. But if your compiler provides auto_ptr,
and you're not worried about portability, it's definitely the way to go.

Miscellanea

I've gotten a couple questions in interviews that I probably should address briefly
here. First, I had at least one applicant ask me why I could do things like

fBar1 = new TBar(*that.fBar1);

when fBar1 is a private member. The answer is that access control in C++ is
done on an class by class basis, not on an instance-by-instance basis. Any instance
of a class can access the private members of any other instance of the same class, so code
like that shown in the example above is legal.

I also sometimes have people write code like

if (fBar1 != 0)
delete fBar1;

This is actually unnecessary. The delete operator in C++ automatically
performs this null check for you, so it's okay to delete a null pointer.

The real answer

Actually, I lied at the beginning of this article. There is an infinitely better
solution to the problem as I stated it than the one we just worked our way to.

To see what I'm getting at, consider two points:

If any other objects contain aliasing pointers to TFoo's TBar objects,
they will be invalid after you assign to the TFoo.

Every object in a well-designed C++ system has a default constructor, a copy
constructor, and an assignment operator.

That's right; you can perform the whole assignment just by calling TBar's
assignment operator. That solution looks like this:

This is so much easier. So why didn't I start with this at the outset? There are
several reasons:

Going with the longer solution was a good way to explore many of the details of C++
memory management and syntax.

The shorter solution won't work if NULL is a valid value for fBar1 and
fBar2. In this case, you'd have to use a hybrid solution: follow the longer
example when you're copying an object into a null pointer, and the shorter example when
you're copying an object into an object. (I'll leave this example as an exercise for the
reader.)

The shorter solution won't work if we want to transactionize the whole assignment. In
other words, if we have a situation where both assignments have to complete
successfully in order for our object to be in a consistent state, we can't use the simple
solution because there's no way to roll back our changes to fBar1 if assigning to
fBar2 throws an exception. You'd have to use the longer example. However, if the
two data members are unrelated, or you know their assignment operators can't throw an
exception, this solution is perfectly adequate.

The shorter solution won't work if TBar is a polymorphic class. Let's take a
closer look at this situation.

Handling polymorphism

I stated in the original problem that you could assume for the purposes of the exercise
that TBar was a monomorphic class that is, that TBar has no
subclasses. If TBar is a polymorphic classthat is, if we know it has or can
have subclassesthen the shorter solution won't work.

Let's pretend for a moment that TBar is an abstract class and that it has two
concrete subclasses, TDerivedBar1 and TDerivedBar2 (yeah, I know I'm
really creative with these names). All we have in our short example are pointers to TBar.
Each assignment above will call TBar'soperator= function, not
the operator= function of TDerivedBar1 or TDerivedBar2
(remember, operator=() isn't virtual). This means that any data members defined
by the TDerivedBar classes won't be copied. This is called slicing, and
it's something you have to watch out for in C++. You must always pay special attention to
whether a class is polymorphic or monomorphic. Polymorphism imposes special restrictions
on what you can do with your objects, and the compiler doesn't enforce these restrictions.

Of course, we could theoretically get around these problems by making TBar's operator=
function virtual. If this->fBar1 is an instance of TDerivedBar1,
you'll call TDerivedBar1::operator=() instead of TBar::operator=(). But
if it's a TDerivedBar2, you're in trouble. TDerivedBar1 isn't going to
know what to do with TDerivedBar2's members; it has nowhere to put them. You
really want it to look like the object pointed to by this->fBar1 has morphed
from a TDerivedBar2 to a TDerivedBar1. There's only one way to do this,
and that's to delete the TDerivedBar2 and new up a brand-new TDerivedBar1.

So our longer solution, where we delete the old objects and replace them with
newly-created objects (either directly or with auto_ptr), is closer to the mark.
But it won't work as written, either. Consider the line

fBar1 = new TBar(*that.fBar1);

If TBar is an abstract class, this will generate a compile-time error, because
you're trying to create an instance of a class that can't be instantiated. If TBar
is a concrete, but polymorphic, class, it'll compile, but you'll get slicing again: the new
expression will only return an instance of TBar, even if the original object was
a TDerivedBar1. At best this isn't a real copy and at worst it has incomplete and
inconsistent state.

The C++ language doesn't provide a built-in way around this problem,
unfortunately if you need to copy an object polymorphically, you have to do it
yourself. The typical way to do this is to define a virtual function called clone()
and have every class that inherits clone() override it to call its own copy
constructor.

Counted pointers and other esoterica

There are a few other points about owning and aliasing pointers that I wanted to make.

The first is that aliasing pointers can be fragile. If you have other objects that have
aliasing pointers to a TFoo's fBar1 object, they'll be bad after a call
to TFoo's assignment operator. This means you have to be careful that any aliases
have a shorter lifetime than the object's owning pointer. Typically, aliases are
short-lifetime objects: You obtain a aliasing pointer to a TFoo's fBar1
at the top of a function, use it to perform some operations on the TBar, and then
let it go out of scope at the end of the function. If you were instead trying to keep the
alias around for an indefinite period of time by making it a data member of some other
object (you'd probably do this for performance or convenience), you're asking for trouble.
The solution to this problem is almost always to make the persistent pointer point to
something with a longer and more well-defined lifespan (for example, the TFoo
that owns the TBar you're interested in).

Of course, there are situations where the relative lifetimes of a network of objects
aren't clear, and therefore it isn't clear who should own whom, or which pointers can be
kept around persistently. The solution to this problem is the same as the solution to the
problem of spending too much time copying objects to maintain clear ownership semantics:
reference counting.

In reference counting, each object keeps track of how many other objects point to it.
These objects refer to the counted object through a counted pointer, which
functions more or less like an aliasing pointer except that it notifies the object when it
no longer refers to it (either because it now points to something else, or because it's
going away). Nobody deletes the counted object; the counted object deletes itself
when no one is left pointing to it.

The C++ standard doesn't provide canned auto_ptr-style objects that do
reference counting, but most frameworks do. It can be expensive, especially in
virtual-memory environments, though, so think twice before resorting to this solution.

An additional caveat is that if you use reference counting to improve performance in a
situation where you'd otherwise be copying objects at the drop of a hat, you don't want to
change the semantics of what you're doing. When you're sharing to avoid unnecessary
copying, every object still expects to behave like he has his own copy of the shared
object. That means the object can only be shared until someone tries to change it.
At that point, the object trying to change it must create a copy of his own. This behavior
is called copy-on-write semantics. Often, it's convenient to create a special
counted-pointer class that enforces copy-on-write semantics transparently. This is almost
never provided by a framework, however; you'd have to code it yourself.

Note, by the way, that in a garbage-collected environment like Java, much of this
material is still relevant. It's true that the garbage collector frees you from having to
worry about deleting things (much as reference counting does, but usually with less
overhead), but that's it. It doesn't free you from having to make sure your object remains
in a consistent state if the assignment fails halfway through. It doesn't free you from
worrying about ownership semantics with resources other than memory blocks (such as disk
files, window-system elements, or locks). And it doesn't free you from having to worry
about when you need to duplicate an object and when an object can be shared. In fact, it
exacerbates the last two problems. In C++, for example, an object can release its system
resources in its destructor, so that all you have to worry about is deleting the object
itself (and not even that if it's in a local variable), whereas you have to explicitly
release them in Java. In C++, you can usually use "const" in interfaces to tell
whether an object can change another object's state or not, where Java lacks
"const," forcing the use of ad-hoc conventions similar to the ones we use in C++
for ownership semantics (or forcing a lot of extra copying). And in both languages, you
have to worry about which objects referring to a shared object should see the change if
the shared object changes (i.e., when do you give an object its own copy of another
object, and when can you legitimately share?)

Garbage collection is definitely not a panacea.

Conclusion

Well, there you have it: a whirlwind tour of some of the finer points of C++ memory
management, as seen through the lens of a simple assignment operator. C++ is a monster of
a language, and writing good code in it takes some practice and some concentration. I hope
these pointers help illuminate some of the more esoteric but important areas of C++
programming for you and help you to write better code.

And who knows? My team still has an opening.

Bibliography

Goldsmith, David, Taligent's Guide to Designing Programs. Reading, MA:
Addison-Wesley, 1994. Affectionately known as "Miss Manners" around Taligent,
this book is the definitive reference for low-level C++ details like the ones we've been
discussing. Most of this material comes straight from this book. Every C++ programmer
should read it.

*The more advanced programmers reading this will
probably be able to come up with one or two better ways of doing this than this particular
example. Stick with me; I'll get to them. I present this example as a good general
solution to the problem that works with almost all C++ compilers.